Understanding statement: F free of rank r $\Rightarrow$ F/F' free abelian of rank r

Question

I found the following exercise on the 12th page of Presentations of Groups:

8. Let $F$ be free of rank $r$. Show that $F/F'$ is free abelian of rank $r$.

First of all, $F'$ is not defined. I'm assuming it stands for a general normal subgroup of $F$.

So, what seems very strange here is that $F$ is not supposed to be abelian, but if I prove that $F/F'$ has the same rank that $F$ has, then, $F/F'$ and $F$ are isomorphic, right? But how can a non-necessarily abelian group be isomorphic to an abelian one?

It seems that someone has already posted a question here related to this: Let free group, F and F/F' have same rank.

But the answers don't address the issue I'm pointing out here.

Answer 1

$F'$ denotes the derived subgroup of $F$, which is just another name for the commutator subgroup $[F,F]$. Thus $F/F'$ is just the abelianization of $F$. This should help you understand the answers in the other thread, or alternatively the answers to the question Abelianization of free group is the free abelian group.